course phy 121 òzòÌÕ½Éݒή´ß¨‚eöæïܶ™êÊassignment #021
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17:35:26 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> horiz v = 12 m/s vert v0 = 0 vert ds = 3m First find vert vf in order to find dt. vf^2 = v0^2 + 2a*ds vf^2 = 0 + 2 (9.8m/s^2) * 3m vf^2 = 19.6 * 3 vf^2 = 58.8 vert vf = 7.67m/s vf = v0 + a*dt 7.67m/s = 0 + 9.8m/s^2*dt dt = .783s Now use dt to find horiz ds assume horiz v is constant vAve = ds/dt 12m/s = ds / .783s ds = 9.396 = 9.4m Use pythagorean theorem to find magnitude vert distance = y = 3m horiz distance = x = 9.4m y^2 + x^2 = mag ^ 2 3^2 + 9.4^2 = mag ^2 9 + 88.36 = 97.36 = mag ^2 magnitude = 9.87 direction = arctan(y/x) = arctan (3 / 9.4) = 17.7 degrees confidence assessment: 2
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17:38:41 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> I knew magnitude and direction of the distance wasn't what was being asked, but I was on such a roll, I ignored my brain telling me that. Of course we were talking about the velocities. I also understand why the 7.7 is negative and the all of the remaining stuff. self critique assessment: 3
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17:46:46 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> AFter I drew my picture I know what I need to find out, but I'm not sure how to do it. First I think I need to figure out how far up the projectile is going to go before it starts back down. Then I figure out how long it will take to get to it's highest point. Then I figure how long it will take to come back down. In order to do all of this I have to determine what part of its initial velocity is horizontal and which part is vertical (that's the part I'm really not sure how to do). I think the horizontal velocity will stay constant, it's the vertical velocity that is going to come to a stop and then accelerate again. confidence assessment: 0
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17:56:04 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> I was right about having to isolate the vertical velocity from the horizontal, but it wasn't as hard as I thought. I see how it was done, but don't know that I could replicate it. self critique assessment: 3
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18:02:48 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> v0x = 20m/s *cos (30degrees) = 17.32 m/s We've already go the dt = 2.7s And I am going to assume that horizontal velocity is constant as in all previous questions of this ilk. therefore ds = vAve* dt ds = 17.32m/s * 2.7s ds = 46.765 = 46.8m confidence assessment: 2
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18:03:16 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> self critique assessment: 3
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